#### Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 8 textbook solution.

Answer : $\frac{\pi}{4}$

Given : $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+e^{x}} d x$

Hint : the formula of  a of $\int_{a}^{b} f(x) d x$ and then add the two equations

Solution : $I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+e^{x}} d x \quad-----(1)$

\begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2}(-x)}{1+e^{-x}} d x \\ &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{1+e^{-x}} d x \end{aligned}

\begin{aligned} &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\ &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{e^{x} \cos ^{2} x}{e^{x}+1} d x \quad-----(2) \end{aligned}

\begin{aligned} &2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos ^{2} x+e^{x} \cos ^{2} x}{1+e^{x}} d x \\ &2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} x d x \end{aligned}

\begin{aligned} &2 I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x\left[\cos 2 x=2 \cos ^{2} x-1\right] \\ &I=\frac{1}{4}\left(x+\frac{\sin 2 x}{2}\right)_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\ &I=\frac{1}{4}\left[\left(\frac{\pi}{2}+0\right)-\left(\frac{-\pi}{2}+0\right)\right] \end{aligned}

\begin{aligned} &I=\frac{1}{4}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\ &I=\frac{\pi}{4} \end{aligned}